All categories

3 years ago

3. Two people push on a car with a force of 5150 N, and the car moves a distance of 8.0m. What is the

Physics

2 answers:

MArishka [77]3 years ago

6 0

I think c not sure tho

tangare [24]3 years ago

3 0

The answer is D, work done is equal to force times it’s distance travelled, so 5150*8.0=41200J

You might be interested in

What is the work done by the friction when the body slides against a rough horizontal surfaces?

katovenus [111]

Negative work

Hope this helps :)

4 0

2 years ago

What is v^2=0.05-4.9 please i need this asap

Margaret [11]

Answer:

v =2.02

Explanation:

v^2=0.05-4.9

v^2=-4.85

square root both side

v=2.02

^^^^this is a not a perfect square

7 0

3 years ago

What is the common formula for work? Assume that W is the work, F is a constant force, 656-06-01-00-00_files/i0060000.jpgv is th

klio [65]

The common formula for work is:
Work = force x displacement
W = F x d

6 0

3 years ago

One of the main factors driving improvements in the cost and complexity of integrated circuits (ICs) is improvements in photolit

nika2105 [10]

Answer:

0.000003782 m

0.000001891 m

0.000001197125 m

Explanation:

$\lambda$ = Wavelength = 248 nm

D = Diameter of beam = 1 cm

f = Focal length = 0.625 cm

The angle is given by

$\theta=\dfrac{1.22\lambda}{D}$

The width is given by

$d=2\theta f\\\Rightarrow d=2\dfrac{1.22\lambda f}{D}\\\Rightarrow d=2\dfrac{1.22\times 248\times 10^{-9}\times 6.25\times 10^{-2}}{1\times 10^{-2}}\\\Rightarrow d=0.000003782\ m$

The required width is 0.000003782 m

Minimum resolvable line separation is given by

$\dfrac{0.000003782}{2}=0.000001891\ m$

The minimum resolvable line separation between adjacent lines is 0.000001891 m

when $\lambda=157\ nm$

$d=2\dfrac{1.22\times 157\times 10^{-9}\times 6.25\times 10^{-2}}{1\times 10^{-2}}\\\Rightarrow d=0.00000239425\ m$

The new minimum resolvable line separation between adjacent lines is

$\dfrac{0.00000239425}{2}=0.000001197125\ m$

6 0

3 years ago

Compare the gravity between these pairs, each consisting of an Earth-like planet and its star. You are given the mass of the pla

Maru [420]

Answer:

The answer is below

Explanation:

Newton's law of gravity states that the force between two bodies is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. The law is expressed by the formula:

$F=G\frac{m_1m_2}{r^2} \\\\Where\ F=force,G=gravitational\ constant, m_1\ and\ m_1=mass\ of\ objects,r\ =distance\ between \ the\ two\ objects.$

The masses and distances for this question is in common units, Therefore the result would be in ratios

a) 4 MEarth / 2 MSolar / 3 AU

The force (F) = (4 * 3) / 3² = 4/3

b) 1 MEarth / 1 MSolar / 1 AU

The force (F) = (1 * 1) / 1² = 1

c) 1 MEarth / 2 MSolar / 2 AU

The force (F) = (1 * 2) / 2² = 1/2

6 0

3 years ago